Question

In the Compton scattering of electrons, by photons incident with wavelength λ,

• A. \(\frac{\Delta \lambda}{\lambda}\) is independent of \(\lambda\)
• B. \(\frac{\Delta \lambda}{\lambda}\) increases with decreasing λ
• C. there is no change in photon's wavelength for all angles of deflection of the photon
• D. \(\frac{\Delta \lambda}{\lambda}\) increases with increasing angle of deflection of the photon

Answer 1

The Correct Option is B, D

To solve this question on Compton scattering, we need to understand how the change in wavelength \(\Delta \lambda\) due to scattering is related to the initial wavelength \(\lambda\) and the angle of deflection.

The Compton wavelength shift equation is given by:

\(\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)\)

where:

• \(h\) is Planck's constant.
• \(m_e\) is the electron rest mass.
• \(c\) is the speed of light.
• \(\theta\) is the angle of deflection.

Notice that the change in wavelength \(\Delta \lambda\) depends on the angle \(\theta\), but not directly on the initial wavelength \(\lambda\). Thus:

1. \(\frac{\Delta \lambda}{\lambda}\) is not independent of \(\lambda\) since \(\Delta \lambda\) itself does not depend on \(\lambda\), but the ratio involves \(\lambda\), which means it can vary.
2. With decreasing \(\lambda\), the same change \(\Delta \lambda\) becomes a larger proportion of the original wavelength, thus \(\frac{\Delta \lambda}{\lambda}\) increases as \(\lambda\) decreases.
3. If \(\theta = 0\), the photon is not deflected, and there is no change in wavelength, hence this option is incorrect as not all angles lead to no change.
4. Finally, \(\Delta \lambda\) increases with increasing angle \(\theta\) because \(\Delta \lambda\) is directly proportional to \(1 - \cos \theta\). Thus, as \(\theta\) increases, \(\Delta \lambda\) and consequently \(\frac{\Delta \lambda}{\lambda}\) also increase.

Based on this analysis, the correct answers are:

• \(\frac{\Delta \lambda}{\lambda}\) increases with decreasing \(\lambda\).
• \(\frac{\Delta \lambda}{\lambda}\) increases with increasing angle of deflection of the photon.